Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2017 Grade 8 Geometry
Problem
Solution
1) What is the maximum number of 3 by 5 by 2 blocks that can be fit into a 16 by 23 by 10 box? The blocks must be stacked with the same orientation.
  1. Find the dimensions of the blocks that divide evenly into dimensions of the box:
        10/5 = 2
        16/2 = 8
  2. The 3rd block dimension, 3, divides into 23 = 7
  3. So the number of blocks that divide into the box are 2x8x7 = 112

2) Triangle ABC has vertices at (2,3), (1, -2), and (-3, 1). The triangle gets reflected across the y-axis, rotated 90 degrees around the origin (counterclockwise ↶ ), then translated down three units. What are the coordinates of the resulting image's vertices?
 Plot the 3 points on this grid.
  1. Reflect across the y-axis
    (this means multiplying y axis values by -1):
        (2,3) = (2,-3)
        (1,-2) = (1,+2)
        (-3,1) = (-3,-1)
  2. Rotate 90 degrees counterclockwise
    (this means exchanging x and y values and negating the y value):
        (2,-3) = (-3,-2)
        (1,+2) = (2,-1)
        (-3,-1) = (1,+3)
  3. Translate 3 units down (subtract 3 from the y value)
        (-3,-2) = (-3,-5)
        (2,-1) = (2,-4)
        (1,3) = (1,0)
  4. New coordinates are (-3,-5),(2,-4),(1,0)

3) What is the measure of angle CDB? Not drawn to scale
  1. The angle at B inside ACDB is supplementary with the 121 degree angle and is 180 - 121 = 59 degrees
  2. The angle at A inside ACDB is vertical with the 85 degree angle and is 85 degrees.
  3. So,
        x + 3x + 59 + 85 = 360
        4x = 216
        x = 54 degrees
  4. Therefore 3x = 162 degrees

Problem
Solution
4) Give the length of side RS in radical form.
 Using the pythagorean theorem:
  • Segment aR = √ 1+1  = √ 2 
  • Segment bR = √ 2+1  = √ 3 
  • Segment cR = √ 3+1  = √ 4 
  • Segment dR = √ 4+1  = √ 5 
  • Segment eR = √ 5+1  = √ 6 
  • Segment RS = √ 6+1  = √ 7 
5) ABCD is a square. A, B, C, and D are the centers of their circles. The two smaller circles are congruent, and the two larger circles are congruent. DE = 4 inches and FA = 2 inches. Find the area of the shaded region. Express the answer to the nearest tenth.
  1. The area of ABCD = (4+2)2 = 36 sq. in.
  2. The 2 circular quarter circles centered at D and B together constitute a half circle of radius 4 inches, so their combined area is:
        A = 42 / 2 = 8
  3. The 2 circular quarter circles centered at A and c together constitute a half circle of radius 2 inches, so their combined area is:
        A = 22 / 2 = 2
  4. The area of the shaded area is the square's area minus these 2 half-circle areas:
        36 - 8 - 2 = 36 - 10 =
        4.6 sq. in.